if a voltage is applied to a capacitor, current flows easily at first and then slows as the capacitor becomes charged. Inductors
behave just the opposite, in that they reluctantly pass current when a voltage is first applied, and then the current passes easily as time passes. If the input voltage is suddenly raised from zero to some constant value, sketch the current in the capacitor, iC, and the inductor, iL, as a function of time. What is the steady-state current in the capacitor and inductor?
1 answer:
Answer: photo!
Explanation:
For the capacitor, the current as a function of time will be
I(t)=V/R e^(-t/RC)
In steady-state, time tends to infinite, so current tends to zero.
For the inductor, the current as a function of time is
I(t)=V/R (1-e^(-tR/L))
In steady-state, time tends to infinite, so current tends to V/R
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Answer:
11.541 mol/min
Explanation:
temperature = 35°C
Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa
note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )
from steam table it is = 5.6291 Kpa
calculate the mole fraction of H ( YH
)
= 5.6291 / 151.95
= 0.03704
calculate the mole fraction of air ( Yair )
= 1 - mole fraction of water
= 1 - 0.03704 = 0.9629
Now to determine the molar flow rate of water vapor in the stream
lets assume N = Total molar flow rate
NH = molar flow rate of water
Nair = molar flow rate of air = 300 moles /min
note : Yair * n = Nair
therefore n = 300 / 0.9629 = 311.541 moles /min
Molar flowrate of water
= n - Nair
= 311.541 - 300 = 11.541 mol/min
Answer:
No.
Explanation:
= Power rating = 1 W
R = Resistance =
V = Voltage =
Power is given by
So
Hence, the resistor is not operating within its power rating.